Should a Big Inductor be Added to a Hartley Oscillator Design?

[Baluncore]

Isn't the feedback fraction C1/C2? If C1 = C2 then the feedback fraction must be equal to 1.

And yet it oscillates. There are many configurations. The feedback gain will depend on how the resonant circuit is grounded.

You forgot about L1

No I didn't. The resonant circuit or "tank" is L in parallel with C1 and C2 in series. One side of L is grounded. The top of L is connected to the base. Half the AC appears across each capacitor, so it acts like a centre tapped "transformer". The emitter drive to the centre tap is effectively doubled to the base.

Spice files attached. To run the .asc file with LTspice, remove the .txt extension that allows me to attach the ascii file to a post. Do the same with the .plt file to get the same traces and colours I get.

 

[DaveE]

Isn't the feedback fraction C1/C2? If C1 = C2 then the feedback fraction must be equal to 1.

You must consider all of the components in the feedback network, theoretically that's every passive component for this configuration*. However, the resistor values are really big and can be ignored, they will have a negligible effect.

One shortcut you can use for a circuit that you know is supposed to work at a particular frequency (in this case you can guess that it will be the LC tank resonant frequency) is to quickly determine the magnitude of the impedances of the parts and compare them. You don't have to be too accurate, you're just looking for really big differences that may allow you to eliminate some from your thinking about what matters. Similar to the way you may separate DC bias calculations from AC dynamic calculations.

Also, for circuits that operate at resonance, you know that the inductive impedance magnitude(s) are the same as the capacitive impedance magnitudes (that's always the situation at resonance). So, for a circuit like this one, you can very quickly identify that Z_o=√(L/C), the characteristic impedance of the tank (which is the magnitude of the L or C impedance at resonance), is the significant impedance for the whole dynamic circuit.

You will save yourself a lot of effort in the future if you just memorize the resonant frequency and characteristic impedance of LC tanks:

ω_o=1/√(LC) and Z_o=√(L/C), also Q=Z_o/R (series LCR) or Q=R/Z_o (parallel LCR) for lossy circuits (which isn't really applicable, in this case).

*post #25, @Baluncore simulation.

 

[tech99]

Ya that's very odd.

The transistor has no bias applied, so it is cut-off and in Class C. No collector current will flow until oscillation starts, then current will flow on positive half cycles. Such oscillators will not start without an intial kick, which might occur due to the switch-on impulse or due to someone touching the circuit.

 

[Baluncore]

Such oscillators will not start without an intial kick, which might occur due to the switch-on impulse or due to someone touching the circuit.

You are correct. Without C3 to float the base voltage the circuit can still maintain oscillation, but it will not start without an initial current pulse in L, or a high voltage applied to the shorted base, neither of which is going to happen in a real implementation of the circuit.

If it could start without C3, the base-emitter junction would then regulate amplitude by entering reverse breakdown at about -5V on each cycle, which is not good for transistor health.

 

[tech99]

You are correct. Without C3 to float the base voltage the circuit can still maintain oscillation, but it will not start without an initial current pulse in L, or a high voltage applied to the shorted base, neither of which is going to happen in a real implementation of the circuit.

If it could start without C3, the base-emitter junction would then regulate amplitude by entering reverse breakdown at about -5V on each cycle, which is not good for transistor health.

I tend to think that even with corect bias, the device will tend to

When you add C3, the bias works OK and it oscillates as expected.

View attachment 277128

View attachment 277130

How can 35mA flow through he 4.7k emitter resistor with a supply of 9 volts?

 

[DaveE]

How can 35mA flow through he 4.7k emitter resistor with a supply of 9 volts?

It can't. It can, however, flow through C2 which has an impedance magnitude of 160ohms at 2.1MHz. I suspect it's more complex than just C2 though since the node is coupled into the tank. You have to look at all of the possible paths.

 

[Baluncore]

I tend to think that even with corect bias, the device will tend to

Tend to what?

In reality it will not start without C3 to isolate the DC base bias from the broadband short of the tank circuit. Without C3, the base bias resistors are irrelevant.

Oscillation of the tank could be started by momentarily connecting the base (and tank) to the +9V supply.

In a Spice model that can be done by using the 'Dot Initial Condition' command to set the initial inductor current to 25 mA by;
.IC I(L) = 25m

Operation without C3 is moot anyhow. No one would build such a circuit. They might however fail to include the capacitor symbol in a schematic diagram. We don't have the original reference, so there is no telling how it happened.